What is the sum from n=1 to infinity of (1/4)^{n-1} ?

The sum from n=1 to infinity of (1/4)^{n-1} is converges

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sum from n=1 to infinity of (1/4)^{n-1}

n = 1 \sum \infty (\frac{1}{4})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{1}{4})^{n - 1}

Apply Series Ratio Test:

converges

= converges

f (x) = \frac{3}{x}, at x = \frac{1}{16}

\int 2 tan^{3} (2 x) sec^{5} (2 x) d x

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 6 x + 5)

f (x) = ln (sin (x^{2} + 1))

\frac{d}{d x} (x^{3} - x^{2} - x + 1)

What is the sum from n=1 to infinity of (1/4)^{n-1} ?
The sum from n=1 to infinity of (1/4)^{n-1} is converges